Problem: Let $f(x)=\begin{cases} \dfrac{15x}{x^3+3x}&\text{for }x\neq 0 \\\\ k&\text{for }x=0 \end{cases}$ $f$ is continuous for all real numbers. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5$ (Choice B) B $8$ (Choice C) C $0$ (Choice D) D $3$
Explanation: $\dfrac{15x}{x^3+3x}$ is continuous for all real numbers other than $x=0$ which means $f$ is continuous for all real numbers other than $x=0$. In order for $f$ to also be continuous at $x=0$, the following equality must hold: $\lim_{x\to 0}f(x)=f(0)$ Since $f(0)=k$, we will obtain the above equality by letting $k=\lim_{x\to 0}f(x)$. So let's find $\lim_{x\to 0}f(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 0}f(x) \\\\ &=\lim_{x\to 0}\dfrac{15x}{x^3+3x} \gray{\text{This is the rule for }x\neq 0} \\\\ &=\lim_{x\to 0}\dfrac{15\cancel{(x)}}{\cancel{(x)}(x^2+3)} \gray{\text{Factor}} \\\\ &=\lim_{x\to 0}\dfrac{15}{(x^2+3)} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq 0) \\\\ &=\dfrac{15}{(0)^2+3} \gray{\text{Direct substitution}} \\\\ &=5 \end{aligned}$ We obtained that if we set $k=5$, then $\lim_{x\to 0}f(x)=f(0)$, which makes $f$ continuous at $x=0$. Since we already saw that $f$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $k=5$.